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Simple lagrangian problem

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  • #1
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Homework Statement


A uniform thin disk rolls without slipping on a plane and a force is being applied at its center parallel to the plane. Find the lagrangian and thereby the generalized force.


Homework Equations


Lagranges equation.


The Attempt at a Solution


This is my first ever exercise of this kind. We first note that U=0 so the lagrangian is simply L=T. We then want to express the kinetic energy T in terms of coordinates, which contain the constraints of the motion implicitly. Therefore we should use polar coordinates. Correct so far?
We then get:
L = T = ½m([itex]\omega[/itex]2r2) (1)

And now lagranges equation says:

d/dt[dT/dqj'] - dT/dqj = Qj
where Qj is the generalized force. There is for this motion one equation of the above kind - one for theta and one for r.

Should I now just differentiate with respect to r and [itex]\theta[/itex] and make two separate equations of the above kind of which I can find the components of the generalized force?
I just don't get anything very sensible when I differentiate the expression for L above with the two variables, and when my teacher did it I think he just differentiated with respect to x - why is that? Shouldn't you use the generalized coordinates?
 

Answers and Replies

  • #2
gabbagabbahey
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This is my first ever exercise of this kind. We first note that U=0 so the lagrangian is simply L=T. We then want to express the kinetic energy T in terms of coordinates, which contain the constraints of the motion implicitly. Therefore we should use polar coordinates. Correct so far?
We then get:
L = T = ½m([itex]\omega[/itex]2r2) (1)
This is what you would get for a disk that was freely spinning subject to no external forces. You need to include the facts that there is a force being applied to the center of the disk,parallel to the plane (what does that mean for the disk, will it only spin or will it have translational motion as well?) and that the disk rolls without slipping (this should give you some relationship between the motion of the center of the disk, and the angular speed of the disk)
 
  • #3
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oh yes right.

So T=½Mv2 + ½I[itex]\omega[/itex]2 = ½(M[itex]\omega[/itex]2R2 + I[itex]\omega[/itex]2)

The translational plus rotational kinetic energy. Do I need this? For rolling without slipping v=[itex]\omega[/itex]R but I am not sure where to use this.
 
Last edited:
  • #4
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hey man, is the above correct?
 
  • #5
vela
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Yeah, it's correct. You've already used ##v=\omega R## to eliminate ##x## and ##\dot{x}##, so now you're just left with ##\theta## and ##\omega=\dot{\theta}##.
 

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